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January 2000
Chapter 12
Manual of Fisheries Survey Methods II: with periodic updates
Chapter 12: Three Methods for Computing the Volume of a Lake
Clarence M. Taube
Suggested citation :
Taube, Clarence M. 2000. Instructions for winter lake mapping. Chapter 12 in Schneider, James C. (ed.) 2000. Manual of fisheries survey methods II: with periodic updates. Michigan Department of Natural Resources, Fisheries Special Report 25, Ann Arbor.
January 2000
Chapter 12
When working with either Method No. 1 or No. 2, first determine area within each contour by tracing around map contour lines with a planimeter. Start with the shoreline contour and continue to the innermost contour line. The resultant readings will be in square inches (the unit of measure of the planimeter). Then, based on the scale of the field map, convert planimeter readings to values of lake area, either in acres or square feet. For very small ponds it may be desirable to compute areas in square feet, but lake area is commonly expressed in acres.
As an example of calculating lake area, and then areas within consecutive contours, assume that a lake map was drawn on the scale of 1 inch equals 100 feet. Then, 1-square inch of map area (planimeter reading) equals 10,000 square feet (100 × 100) of lake area, or 0.22957 acres (10,000/43,560). Further assume for this example that the lake has a maximum depth of 23 feet, depth contours were drawn for each 5-foot interval, and planimeter readings for the area within the contours were as in the following table:
Planimeter reading Calculated area Depth contour (feet) (square inches) Square feet Acres
(Shoreline contour) 0 210.0 2,100,000 48. 5 150.0 1,500,000 34. 10 110.0 1,100,000 25. 15 83.5 835,000 19. 20 21.7 217,000 5. (Maximum depth) 23
Those calculated areas were obtained by conversion factors of 10,000 for square feet and 0.22957 for acres. Area in acres could have been calculated by dividing area in square feet by 43,560 (number of square feet per acre). Note that 48.2 acres is the calculated total area of the lake.
As an illustration of computing water volumes, we continue to use the sample data given above. By Method No. 1, calculations of volume (in acre-feet) are as follows:
Depth strata Equation Acre feet
0 - 5 ft : 31 ∗^5 (^48.^2 +^34.^4 +^48.^2 *^34.^4 ) = 205.
5 - 10 ft : 31 ∗^5 (^34.^4 +^25.^3 +^34.^4 *^25.^3 ) = 148.
10 - 15 ft : 31 ∗^5 (^25.^3 +^19.^2 +^25.^3 *^19.^2 ) = 110.
15 - 20 ft : 31 ∗^5 (^19.^2 +^5.^0 +^19.^2 *^5.^0 ) = 56.
20 - 23 ft : 31 ∗^3 (^5.^0 ) = 5.
Total volume = 526.
Note that the volume calculation for the lowermost layer (20-23 ft) uses the cone formula: volume = 1/3(HA). By applying the formula here we assume that the maximum depth of 23 feet occurred only in a small area. If the maximum depth of 23 feet prevailed over an extensive area, then encircle this area with a contour line, determine its area with a planimeter, and use the frustum formula to calculate volume of the 20-23-foot zone.
January 2000
Chapter 12
By Method No. 2, example calculations of volume (in acre-feet) are as follows:
Depth strata Equation Acre feet
0 - 5 ft : 21 ∗^5 (^48.^2 +^34.^4 ) = 206.
5 - 10 ft : 21 ∗^5 (^34.^4 +^25.^3 ) = 149.
10 - 15 ft : 21 ∗^5 (^25.^3 +^19.^2 ) = 111.
15 - 20 ft : 21 ∗^5 (^19.^2 +^5.^0 ) = 60.
20 - 23 ft :
3 ( 5. 0 ) 2
Total volume = 535.
When using acres for area values and feet for depth values, volume will be in acre-feet. An acre-foot of water is 1 acre of water 1 foot deep, i.e., 43,560 cubic feet.
In Method No. 3, all soundings of the lake are summed, then divided by the number of soundings to obtain average depth. Lake volume equals average depth times lake area. Lake area is determined by planimeter measurements on the field map, as described above.
12.5 Comparison of the three methods
The three methods give approximate rather than exact volumes of lakes, but these approximations are close enough to the true values for fisheries work. These methods usually give quite similar results, as demonstrated by the three example lakes below:
Computed volumes (acre-feet) Name of lake Area (acres) Location Method No.1 Method No.2 Method No. Frost 60.0 Ogemaw Co. 1,949 1,963 1, Robinson 20.3 Oakland Co. 64 63 58 Eagle 19.9 Oakland Co. 137 142 138
However, based on years of experience, Methods 2 and 3 often give slightly higher values than Method 1 (the Frost Lake example is one of the exceptions). The slight difference appears to be related to lake basin shape, but has no practical significance to fisheries work. Methods Nos. 2 and 3 are preferable to Method No. 1 from the standpoint of simplicity. Assuming that the lake in question is shaped like a series of frustums, the formula of Method No. 1 is mathematically correct.
Written 4/1947 by C. M. Taube Revised 3/1976 by C. M. Taube Slightly revised 1/2000 by J. C. Schneider
